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Life, AI, and the Stars
By
Ian Beardsley
Copyright © 2021 by Ian Beardsley
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Table of Contents
Introduction………………………………..3
Molecular Geometry………………………4
Bone As A Mathematical Construct……….5
Breaking Down Bone……………………..13
Perfect Equations………………………….18
The Ratios Come Together………………..21
The Protoplanetary Disc…………………..23
The Stars…………………………………..26
The Masculine And Feminine…………….30
The Planetary Equations………………….31
Binary Digital Logic……………………..35
Weird Arithmetic…………………………37
Another Scheme………………………….45
Amino Acids……………………………..46
Asymmetry In AI Elements………………48
Fundamental Equations…………………..59
Formulating A Notation…………………..64
Perfect Equations…………………………67
The Damped Harmonic Oscillator……….72
QM and Electrostatics……………………75
The Planetary Equations………………….78
The Geometry…………………………….80
Within A Compound………………………83
Conclusion………………………………..92
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Introduction
In my works The Mathematical Nature of Life (Beardsley 2021) and Perfect Equations Beardsley (2021) I
set out to find if the the elements and compounds characteristic of life and artificial intelligence (AI) do
not just conform to chemical law, but if they are purely mathematical independently of the use of
chemistry to describe them, and if they are connected to one another. The simplest example of this for
biological life and AI would be that the most basic organic compound is HNCO (isocyanic acid) where H
(hydrogen), N (nitrogen), C (carbon), and O (oxygen) are the most abundant biological elements. Indeed
biological elements are for the most part organic, which means they are made of long chains using carbon
with hydrogen, which they can form because C is C4- and H is H+ meaning we can have:
And in isocynaic acid we have:
H-N=C=O
Where H is H+, N is N3-. C is C4-, O is O2-, the H uses its single bond with one from nitrogen, leaving
N2- or two bonds which go to C leaving for it C2- which goes to oxygen that needs it because it is O2-.
Thus all is satisfied by chemical law. In my search for mathematical law, I find it exists in the case of
HNCO and the AI semiconducting element silicon (Si) and its doping agents P and B as such (by molar
mass):
This paper strives to break down such mathematical equations for biological life and artificial intelligence
into their components to find what is acting to create such constructs. In the second book I actually
brought the planets into the mix with some very interesting results. As another example, water and air, the
main physical constituents that interact with life we have:
C + N + O + H
P + B + Si
≈ ϕ
ϕ =
a
b
=
5 − 1
2
c = b + a
a
b
=
b
c
H
2
O
air
≈ ϕ
air = 0.25O
2
+ 0.75N
2
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By molar mass for air as a mixture (not a compound). With this air is 29.0 grams per mole.
Molecular Geometry
We will want to break down our equations into the components of their geometric relationships and see if
they predict the bond angles of some of the basic substances considered. We will look here at linear,
trigonal planar, and tetrahedral.
Linear, like CO2 (carbon dioxide) its bond angle is 180 degrees:
Trigonal planar, like SO3 (sulfur trioxide) its bond angle is 120 degrees:
O
|
S
/. \
O. O
That is, S is at the center and the O atoms are 120 degrees apart due to the even division of 360/120=3.
Tetrahedral, like methane (CH4) one of the the primordial gases that may have contributed to making
some of the amino acids, the building blocks of life as show by Miller and Urey in the early origins of
life:
This is 109.5 degrees apart from arcos (1/3) = 109.5
But what if we are considering not just neutral molecules but polyatomic anions that have a net charge. In
such instances, the free electron pairs compress the expected 120 degree bond angle in the atoms around
the central atom to 115 degrees as with the nitrite ion NO2-:
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Similarily we have for O3 (ozone) that the bond angle is 116 degrees in its deviation from 120 degrees.
The configuration is:
Both of these anions are important to the life and the theory of how life forms. O-zone is more of a
physical component in that in the stratosphere it absorbs UV radiation harmful to life.
Bone As A Mathematical Construct
What better place to begin than with than bone as it is the basic framework around which skeletal life is
structured, the vertebrates. Here is what I found in bone as a mathematical construct:
In my exploration of the connection between biological life and AI the most dynamic component is that of
bone. It affords us the opportunity to look at:
Multiplying Binomials
Completing The Square
The Quadratic Formula
Ratios
Proportions
The Golden Ratio
The Square Root of Two
The Harmonic Mean
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Density of silicon is Si=2.33 grams per cubic centimeter.
Density of germanium is Ge=5.323 grams per cubic centimeter.
Density of hydroxyapatite is HA=3.00 grams per cubic centimeter.
This is
where
Where HA is the mineral component of bone, Si is an AI semiconductor material and Ge is an AI
semiconductor material. This means
The harmonic mean between Si and Ge is HA,…
This is the sextic,…
Which has a solution
Where x=Si, and y=Ge. It works for density and molar mass. It can be solved with the online Wolfram
Alpha computational engine. But,…
3
4
Si +
1
4
G e ≈ H A
H A = Ca
5
(PO
4
)
3
OH
Si
H A
Si +
[
1 −
Si
H A
]
G e = H A
2 SiGe
Si + G e
≈ H A
x
2
(x + y)
4
− x y(x + y)
4
+ 2x y
2
(x + y)
3
− 4x
2
y
2
(x + y)
2
= 0
Si
G e
=
1
2 + 1
1
H A
2
Si
2
−
G e
H A
2
Si +
[
G e
H A
− 1
]
= 0
Si =
1
2
G e
±
H A
G e
H A
2
−
4G e
H A
+ 4
Si = G e − H A
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Si
H A
Si +
[
1 −
Si
H A
]
G e = H A
Si
2
H A
+ G e −
Si
H A
G e ≈ H A
1
H A
Si
2
−
G e
H A
Si + G e ≈ H A
1
H A
2
Si
2
−
G e
H A
2
Si +
G e
H A
≈ 1
1
H A
2
Si
2
−
G e
H A
2
Si +
G e
H A
− 1 ≈ 0
1
H A
2
Si
2
−
G e
H A
2
Si +
[
G e
H A
− 1
]
= 0
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We see that the square of the binomial is a quadratic where the third term is the square of one half the
middle coefficient. This gives us a method to solve quadratics called completing the square:
(x + a)(x + a) = x
2
+ 2a x + a
2
(x + a)
2
= x
2
+ 2a x + a
2
a x
2
+ bx + c = 0
a x
2
+ bx = − c
x
2
+
b
a
x = −
c
a
(
1
2
b
a
)
2
=
1
4
b
2
a
2
x
2
+
b
a
x +
1
4
b
2
a
2
= −
c
a
+
1
4
b
2
a
2
(
x +
1
2
b
a
)
2
=
b
2
− 4a c
4a
2
x +
b
2a
=
±
b
2
− 4a c
2a
x =
−b
±
b
2
− 4a c
2a
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1
H A
2
Si
2
−
G e
H A
2
Si +
[
G e
H A
− 1
]
= 0
x =
−b
±
b
2
− 4a c
2a
a =
a
H A
2
b = −
G e
H A
2
c =
[
G e
H A
− 1
]
b
2
− 4a c =
G e
2
H A
4
− 4
1
H A
2
[
G e
H A
− 1
]
=
G e
2
H A
4
−
4G e
H A
3
+
4
H A
2
=
1
H A
2
[
G e
2
H A
2
−
4G e
H A
+ 4
]
b
2
− 4a c =
1
H A
(
G e
H A
− 2
)
2
x =
Ge
HA
2
±
1
HA
[
Ge
HA
− 2
]
2
HA
2
=
1
2
G e
±
1
2
H A
[
G e
H A
− 2
]
=
1
2
G e
±
1
2
G e − H A
Si =
1
2
G e +
1
2
G e − H A
Si = G e − H A
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Si ≈ G e − H A
H A ≈
2 SiGe
Si + G e
Si ≈ G e −
2 SiGe
Si + G e
(Si + G e)Ge
Si + G e
−
(Si + G e)Si
Si + G e
−
2 SiGe
Si + G e
= 0
G e
2
− 2 SiGe − Si
2
Si + G e
= 0
x
2
− 2x y − y
2
= 0
x
2
− 2x y = y
2
x
2
− 2x y + y
2
= 2y
2
(x − y)
2
= 2y
2
x − y =
±
2y
x = y + 2y
x = y(1 + 2)
x
y
= 1 + 2
y
x
=
1
2 + 1
Si
G e
≈
1
2 + 1
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A ratio is and a proportion is which means a is to b as b is to c.
The Golden Ratio
and.
or
a
b
a
b
=
b
c
(
Φ
)
a
b
=
b
c
a = b + c
a c = b
2
c =
b
2
a
a = b +
b
2
a
b
2
a
− a + b = 0
b
2
a
2
− 1 +
b
a
= 0
(
b
a
)
2
+
b
a
− 1 = 0
(
b
a
)
2
+
b
a
+
1
4
= 1 +
1
4
(
b
a
+
1
2
)
2
=
5
4
b
a
= −
1
2
±
5
2
b
a
=
5 − 1
2
a
b
=
5 + 1
2
ϕ =
5 − 1
2
Φ =
5 + 1
2
ϕ =
1
Φ
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The mineral component of bone hydroxyapatite (HA) is
The organic component of bone is collagen which is
We have
%
Ca
5
(PO
4
)
3
OH = 502.32
g
m ol
C
57
H
91
N
19
O
16
= 1298.67
g
m ol
Ca
5
(PO
4
)
3
OH
C
57
H
91
N
19
O
16
= 0.386795722
ϕ = 0.618033989
1 − ϕ = 0.381966011
Ca
5
(PO
4
)
3
OH
C
57
H
91
N
19
O
16
≈ (1 − ϕ)
0.381966011
0.386795722
100 = 98.75
Si
G e
=
28.09
72.61
= 0.386861314 ≈ (1 − ϕ)
Si
G e
≈
Ca
5
(PO
4
)
3
OH
C
57
H
91
N
19
O
16
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Breaking Down Bone
Essentially, in our mathematical formulation of bone, we had that
Which resulted in that the AI elements:
By way of the mineral component of bone HA (hydroxyapatite) is the harmonic mean between Silicon
and Germanium the primary semiconductor elements, which are really the skeleton on AI. Thus, we need
to break down the harmonic mean between Si and Ge into its geometric representation, and through find
what its components are if we are to get any sense of the dynamics. Here I do that in the following
illustration:
Si ≈ G e − H A
H A ≈
2 SiGe
Si + G e
Si
G e
≈
1
2 + 1
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We see that through bone Si and Ge predict an angle of about 116 degrees. This is not the case of linear at
180 degrees, or tetrahedral pyramidal at 109.5 degrees, but is the instance of trigonal planar, but not of
neutral molecules, which is 120 degrees, but of trigonal planar for polyatomic anions such as the nitrite
ion:
And O-zone (Not an anion but has free electrons due to a single bond):
We also will want to look at that aspect of the other mathematical constructions we found for bone:
Which means:
Ca
5
(PO
4
)
3
OH
C
57
H
91
N
19
O
16
≈ (1 − ϕ)
Si
G e
=
28.09
72.61
= 0.386861314 ≈ (1 − ϕ)
Si
G e
≈
Ca
5
(PO
4
)
3
OH
C
57
H
91
N
19
O
16
Si
G e
=
1
2 + 1
2 − 1 ≈ 1 − ϕ
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Which means we want to explore this as well:
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The thing that is important here is that both and being irratiational, are the operative components to
And this is where we come to some sort of a conclusion as to how square root of two and the golden ratio
in molar mass and density of bone might determine something physical about Nature. We consider that
such physical proportions in the lattice that makes bone, carry through to the proportions found in
humans:
Where we find the square root of two is interesting; we are all familiar with the golden ratio phi in the
human body. For instance in the height divided by the distance from the bottom of the feet to the navel.
But, the answer comes from archaeology.
The intermembral index compares the forelimbs of vertebrates to their hindlimbs. A ratio greater than one
means the forelimbs are longer than the hindlimbs and less than one the hindlimbs are longer. It is this
ratio that tells paleontologists a great deal about the manner of propulsion of a vertebrate.
The chimpanzee index is 106, or 1.06 in other words as a fraction, meaning their forelimbs are longer
than their hindlimbs compared to humans, which are around 68-70 or 0.68 to 0.7 meaning their hindlimbs
are longer than their forelimbs. Thus we see they have their forelimbs are longer for climbing, arm
hanging and swinging activities. The longer hindlimb of humans means they depend sole on these for
propulsion in bipedal walking. Lucy, the 3.2 million year old hominid (Australopithecus Afarensis) has
index 88 (0.88) intermediate between humans and chimpanzees, and this due to a shortened humerus, not
elongated thigh, showing arm length reduced first in the evolutionary trend toward being bipedal. She
probably used hindlimb for bipedal propulsion and forelimbs for climbing.
Measuring myself I find I have humerus+radius=22”, and femur+tibia=32”. My intermembral index is
about i=22/32=0.6875. And here is our :
Are we evolving towards an intermembral index of ?!
ϕ
2
2 − 1 ≈ 1 − ϕ
2
1
i
=
1
0.7
= 1.42857 ≈ 2 = 1.414
2
of 18 93
Perfect Equations
I made two equation sets for the planetary orbits one in terms of core AI semiconducting elements Si and
Ge and one in terms of the mathematical constants and Euler’s number e. The result was that for the
first set of equations the equation for Venus was perfect and for the second set of equations the equation
for Mars was perfect. These are:
1AU= average earth-sun separation. This is interesting because Venus and Mars have always been of
great interest to space programs: they are solid and the closest to us. The Russians focused their space
program on Venus sending several probes there, presumably because it is our sister planet (close to the
same as the Earth in size and mass) to understand why she underwent a runaway greenhouse effect
presumably so we can understand how to prevent one here. And the United States has sent several rovers
to Mars to search for life and to better understand the planet in our solar system that can be colonized.
Interestingly, in 2020 we have discovered evidence of microbial life in the Venus atmosphere.
If we want to explore the geometry of the Mars equation we can express geometrically, but how do we
do as such for Euler’s number e? I propose by looking at because is the ratio between the side of
an equilateral triangle to its radius and is while (e=2.718…). Also, in the Venus
equation we have venus=0.72 AU. Thus all seems to come together with this approach. Thus we have:
Where denotes Mars as the fourth planet. We have:
Using P4=1.52, phi=0.618, Ge=72.64, and Si=28.09 we have:
ϕ
ve n us =
1
G e
2
2 SiGe +
Si
3
Ge
1 +
Si
2
Ge
2
= 0.72AU
m ars = ϕ
2
e
2−ϕ
= 1.52AU
ϕ
3
1
3
3 ≈ 1.72
e ≈ 2.72
1
G e
2
2 SiGe +
Si
3
Ge
1 +
Si
2
Ge
2
≈ ( 3 − 1)AU
ϕ
2
(1 + 3)
2−ϕ
= P
4
P
4
ln
[
P
4
ϕ
2
]
= (2 − ϕ)l n
2 Si
G e
(
Si
2
Ge
2
+ 1
)
+
Si
3
G e
3
(
Si
2
Ge
2
+ 1
)
+ 2
of 19 93
This gives:
1.38=1.2967 is
Is about 94% accuracy. We see that the common geometry to Mars and Venus is . Square root three is
irrational like in , and . This is what lends them their properties. This is clear in my illustration
on the next page of the Vesica Piscis. The interesting thing about the square root of two is that it has the
property:
One is to the square root of two as the square root of 2 is to two.
ln(3.9798) = 1.382l n
(
56.18
72.64(1.495)
+
22164.36
383290.0(1.495)
+ 2
)
1.2967
1.38
100 = 0.9396
3
5
ϕ
2
1
2
=
2
2
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of 21 93
The Ratios Come Together
Let us say a/b=x, the golden ratio. Then,…
Let us differentiate this implicitly:
Which is similar to Euler’s number, e because it is the base such that is itself :
But
Which says for this angle the x-component equals the y-component is that is , x=1/2 bisects a right
angle. Which similar in concept to Euler’s number e because it is the base such that is itself . But
if , then:
It is the diagonal of the unit square. We notice something interesting happens:
, ,
Where is the cosine of 30 degrees, in the unit equilateral triangle in which the altitude has been drawn
in (fig 14):
ϕ =
b
a
=
5 − 1
2
x
2
− x − 1 = 0
d
dx
x
2
−
d
dx
x −
d
dx
1 = 0
2x − 1 = 0
x =
1
2
d
dx
e
x
e
x
d
dx
e
x
= e
x
sin 45
∘
= cos45
∘
=
2
2
1
2
90
∘
d
dx
e
x
e
x
sin 45
∘
= cos45
∘
=
2
2
2cos
π
4
= 2
2cos
π
n
=
2cos
π
4
= 2
2cos
π
5
= Φ
2cos
π
6
= 3
3
of 22 93
Since air is 25% N2 and 75% O2, the molar mass of air as a mixture is 29.0 g/mol. I found air over H2O
is the golden ratio, and in total that:
Where ZnSe is zinc selenide, an intrinsic semiconductor, intrinsic meaning that it does not have to be
doped to semiconduct.
air
H
2
O
≈ Φ
0.75N
2
+ 0.25O
2
≈ a ir
−2cos
(
π
4
)
+ 2cos
(
π
5
)
+ 2cos
(
π
6
)(
Z n
Se
)
≈
air
H
2
O
2cos
(
π
4
)
= 2, 2cos
(
π
5
)
= Φ, 2cos
(
π
6
)
= 3
of 23 93
The Protoplanetary Disc
But, why describe the orbits of the planets in terms of AI Semiconducting elements? My answer is to do
something cosmic: there is great satisfaction in finding the connection between two things that seem
universes apart. And here I present a reason for looking at such a thing, by considering the protoplanetary
disc from which the planets formed. First we form a table of the masses of the planets.
of 24 93
Taking the protoplanetary disc as a thin disc we integrate from its center to the edge, with density
decreasing linearly to zero at the edge. Thus, if the density function is given by
And, our integral is
The mass of the solar system adding up all the planets yields
That accounts for
82% of the mass of the solar system not including the sun, that is, of the protoplanetary disc
surrounding the sun.
Using germanium alone, we get,
If we weight the mixture of silicon and germanium as 1/3 and 2/3, then we have
Which is very close.
93%
This is all very good, because I only used the planets and asteroids.
Weighting silicon and germanium as 1/4 and 3/4 we have
Si + G e
2
=
2.33 + 5.323
2
= 3.8265g /c m
3
ρ(r) = ρ
0
(
1 −
r
R
)
M =
∫
2π
0
∫
R
0
ρ
0
(
1 −
r
R
)
r d r d θ
M =
πρ
0
R
2
3
π (3.8265)(7.4 × 10
14
)
2
3
= 2.194 × 10
30
gr a m s
M = 2.668 × 10
30
gr a m s
2.194
2.668
100 =
π (5.323)(7.4 × 10
14
)
2
3
= 3.05 × 10
30
gr a m s
π (4.32467)(7.4 × 10
14
)
2
3
= 2.48 × 10
30
gr a m s
2.48
2.668
100 =
of 25 93
Which accounts for
98%
Of the mass of the solar system (very accurate).
This mixture of 1/4 to 3/4 is a combination that exists in the Earth atmosphere which is approximately the
mixture of oxygen to nitrogen. The earth atmosphere can be considered a mixture of chiefly O2 and N2 in
these proportions:
Air is about 25% oxygen gas (O2) by volume and 75% nitrogen gas (N2) by volume meaning the molar
mass of air as a mixture is:
By molar mass the ratio of air to H20 (water) is about the golden ratio:
I am not saying the solar system was a thin disk with density of the weighted mean somewhere between
silicon and germanium, but that it can be modeled as such, though if the protoplanetary disk that eclipses
epsilon aurigae every 27 years is any indication of what a protoplanetary cloud is like, it is a thin disk in
the sense that it is about 1 AU thick and 10 AU in diameter. This around a star orbiting another star.
π (4.4 . 57475)(7.4 × 10
14
)
2
3
= 2.623 × 10
30
gr a m s
2.623
2.668
100 =
0.25O
2
+ 0.75N
2
≈ a ir
air
H
2
O
≈ Φ
of 26 93
The Stars
Not only is the Venus equation perfect in the equation set in terms of AI elements, but in the Mars
equation in terms of and in the equation set in terms of these constants. We have:
Let us solve for x in:
1.52=0.72x
x=2.11111=19/9
This is very close to fluorine (F) over beryllium (Be):
Beryllium (see illustrations) is pivotal to the production of carbon, which is in turn the core element to
biological life compounds. Stars produce carbon by combining two helium (He) atoms to make
Beryllium. The Beryllium then combines with another helium atom to make carbon. Beryllium rarely
occurs in Nature because it is usually depleted in the reaction in stars using it to make heavier elements. I
found a connection of beryllium to carbon by way of silicon. The radius of silicon is Si=0.118nm. If we
say it is inscribed in a dodecagon (12 sided) regular polygon of side 1, then carbon inscribes in a regular
octagon (8 sided) of side 1. This is the eight of The beryllium-8 (four protons, four neutrons) that makes
carbon by combining with helium.!
ϕ
e
ve n us =
1
G e
2
2 SiGe +
Si
3
Ge
1 +
Si
2
Ge
2
= 0.72AU
Mars = ϕ
2
e
2−ϕ
= 1.52AU
F
Be
=
19.00g /m ol
9.01g /m ol
= 2.108768 ≈ 2.1
of 27 93
!
of 28 93
"
of 29 93
Flourine is the most electronegative element making it highly reactive. It reacts with everything but
argon, neon, and helium. Flourine combines with carbon to make fluorocarbons. The primary mineral
source is fluorite and is mined for steelmaking which turn results in byproducts used aluminum refining.
Its electron configuration is , giving it seven outer electrons so that it needs one more to be
filled, to give it eight outer electrons because it tends to capture an electron to make it isoelectronic with
the noble, or inert gas neon.
Thus we have the following Equation that through the planets Venus and Mars, the solid planets directly
on either side of the Earth, relates artificial intelligence to biological life:
We have that at the beginning of the Universe hydrogen and helium were created. Then the stars formed
and synthesized these into the heavier elements. I find if we include in the category of life not just the
biological elements, but the AI elements, we can find a mathematical equation for a pattern in the periodic
table of the elements that predicts the synthesis of such elements in stars. For instance, Beryllium 8 plus
helium 4 synthesizes to make the biological core element carbon C. Magnesium plus helium 4
synthesized to make the core AI element silicon Si. If we say that Element 4 is Beryllium Be and write it
, and helium He is element 2 and write it , and use this convention for all of the elements, we have
for the production of these elements by stars, and their molar masses in the periodic table the following
equation:
1S
2
2 S
2
P
5
1
G e
2
2 SiGe +
Si
3
Ge
1 +
Si
2
Ge
2
F
Be
= ϕ
2
e
(2 −ϕ)
E
4
E
2
E
2n−2
+ E
2
= E
2i−2
= (4k + 4)g /m ol
n = (3,4, 5,6, …)
i = (4,5, 6,7, …)
k = (2,3, 4,5, …)
of 30 93
The Masculine and Feminine
Here I will suggest the term masculine silicon and feminine germanium in place of positive (p-type
silicon) and negative (n-type germanium) respectively. And, I will denote them , and , which are dagger
and double dagger.
We say since silicon (Si) doped with boron (B) is p-type silicon because boron being in group 13 only has
three valence electrons and silicon wants four, giving it a deficiency of negative electrons and thus a net
positive distribution that can carry electrons, holes they can fall into. Thus I will say:
And since we say germanium (Ge) doped with phosphorus (P) is n-type silicon because phosphorus being
in group 15 has five valence electrons and germanium, being in group 14 like silicon, wants four
electrons. Thus it has a surplus of negative electrons and thus a net negative distribution that can carry a
current. Thus I will say:
Since B/Si=10.81/28.09=0.3867 and Ge/P=72.64/30.97=2.345 we have:
and.
Now we turn from this construct of the masculine and feminine in AI to the masculine and feminine in
biology.
We consider the female sex hormone estradiol (estrogen , E):
And the male sex hormone testosterone (T):
And, cholesterol (Ch) from which both are made:
And notice,…
And we consider the semiconductor materials used to make AI:
†
‡
B
Si
= †
G e
P
= ‡
† = 0.3867
‡ = 2.345
C
18
H
24
O
2
= 272.38g /m ol
C
19
H
28
O
2
= 288.42g /m ol
C
27
H
46
O = 386.65g /m ol
Ch + T
E
= 2.5
G e
Si
= 2.6
of 31 93
And write,…
We notice that the masculine (T) is in inverse relation to the feminine (E), but that the two add up to on
whole (Ch) in that the masculine has coefficient 1-Si/Ge and the feminine has coefficient 1-Ge/Si. This
expresses the inverse relationships between man and woman.
I interpret this as the masculine (T) is in inverse relation to the feminine (E), but that the two add up to a
whole (Ch) in that the masculine has coefficient 1-Si/Ge and the feminine has coefficient 1-Ge/Si that is
they are inverse relation but compliment one another. How would an AI use this information to determine
its sex?…
The male is reduced less in the difference between 1 and Si/Ge, but the the female is reduced less by
having Ge in the numerator. It is really quite egalitarian.
We now see that:
And this shows the connection of masculine and feminine AI to masculine and feminine biological life.
The Planetary Equations
Though I did not derive these equations, but guessed at them, it was an educated guess which proceeded
from the argument, that the first planet being the closest to the Sun, sets the idea in motion, that idea being
that its distance is in the simplest expression between Si, and Ge, possible; the ratio between them. Thus
for Mercury ( ) we have
In astronomical units because we take an astronomical unit to be 1 AU at earth orbit, because the earth is
the one planets in the solar system that is highly hospitable to life.
To make our guess at the distance of the next planet, we first guess at the simplest idea r(n)=n. Seeing this
does not work we go the next simplest expression r(n)=2n. Seeing this does not work we go one step
higher in complexity and try . We see this does not work either so we go one step of complexity
beyond that and we try . And, we find this works. But it must be tempered with a factor of 0.3
and and adjusted by 0.4. Thus we have the Titius-Bode Rule for the distribution of the planets:
Ch + T
E
=
G e
Si
T =
G e
Si
E − Ch
E =
Si
G e
(T + Ch)
T
(
1 −
Si
G e
)
+ E
(
1 −
G e
Si
)
= Ch
(
Si
G e
− 1
)
T
(
1 − †
)
+ E
(
1 − ‡
)
= Ch
(
Si
G e
− 1
)
P
1
P
1
=
Si
Ge
r (n) = n
2
r (n) = 2
n
of 32 93
To get the next equation for . Since we are dealing with a doubling effect we guess it is ,
or . Which is close but a little too high. So we guess at something lower and that it involves twice
their product and something big, like the sum of their squares to reduce the number to prevent the product
from being too large. We guess that that value then is . This is a little too low so we average
the two to get exactly the result we need:
Since the next planet, our Earth, must be at a greater value than Venus, and the uppercase scenario for
Venus is , we want to reduce 2SiGe by an amount less than the sum of their squares, we
reduce it by the difference of their squares and get for Earth:
It takes a bit of doing, but the next planet ( ) is:
The next location is the asteroids from 2.2 AU-3.2 AU. Since Mars is the last terrestrial planet (solid), and
the asteroid is a bunch of rocks that could not form into a solid planet and after the asteroids we have the
gas giants Jupiter, Saturn, Uranus, Neptune…and this represents a flipping around point around the
asteroid belt, I started counting again with a new pattern starting with Jupiter as , and flipped the Earth
equation and turned a minus sign to a plus to obtain the Jupiter equation, which is a quadratic in its
simplest form in the numerator, and a product in its simplest form in the denominator, which is great
because it is the first planet after the asteroid belt and is hence:
Now what happens is that a simple pattern forms. We simply multiply Jupiter by 2 to get Saturn, then by
four to get Uranus, and finally by six to get Neptune. We have:
r (n) = 0.4 + (0.3)2
n
n = − ∞, 0,1, 2,…
P
2
= Venu s
2P
1
2
Si
Ge
2SiG e
Si
2
+ G e
2
P
2
=
1
Ge
2
2SiG e +
Si
3
Ge
1 +
Si
2
Ge
2
2SiG e
Si
2
+ G e
2
P
3
=
2SiG e
Ge
2
− Si
2
P
4
= Mars
P
4
=
2SiG e
2
(Si − Ge)
2
(Si + G e)
P
1
P
1
=
Si
2
+ 2SiG e + G e
2
SiG e
P
2
= Sat ur n =
2(Si + G e
2
SiG e
of 33 93
Thus we have the following table:
P
3
= Ura nu s =
4(Si + G e
2
SiG e
P
4
= Nept u n e =
6(Si + G e
2
SiG e
of 34 93
of 35 93
The Planets in Binary Digital Logic
I have devised a scheme for the planets in terms of the golden ratio conjugate phi and Euler’s number
e:
So that for the planets exterior to the asteroid belt or where
is is the asteroid belt ( ). .
Which is the solution to the differential equation
Where have we seen this? In computer science.
means
Where n is the number of bits in a number N in binary. We write in binary
0=0
1=1
10=2
11=3
100=4
101=5
110=6
(ϕ)
(1 − ϕ)e
ϕ
= 0.7AU = Venu s
ϕe
(1−ϕ)
= 0.9AU = Ear th
ϕ
2
e
(2−ϕ)
= 1.52 = Mars
2ϕe
(2−ϕ)
= 4.9 = Jupiter
4ϕe
(2−ϕ)
= 10 = Sat ur n
8ϕe
(2−ϕ)
= 19.69 = Ur a nus
16ϕe
(2−ϕ)
= 39.38 = Nept un e
P
n
= 2
n
ϕe
(2−ϕ)
P
n
= c2
n
c = ϕe
(2−ϕ)
= 2.461
c2
0
= c = 2.461
P
0
ϕ =
5 − 1
2
P
1
= Jupiter
P
2
= Sat ur n
P
3
= Uranus
P
4
= Neptu ne
d
2
y
dn
2
− 2log(2)
dy
dn
+ log
2
(2)y = 0
log
2
N = n
2
n
= N
of 36 93
111=7
1000=8
1001=9
1010=10
1011=11
1100=12
1101=13
1110=14
1111=15
10000=16…
But what is interesting about this?
You can’t have a fractional number of bits, thus the spectrum is quantized according to whole
number solutions of
But so are the planets given by
Meaning, since we have 2, 4, 8. 16 that the planets are quantized into whole number orbits according to
computer binary with Jupiter as 2, Saturn as 4, Uranus as 8, and Neptune as 16 if we do it in terms of
Euler’s number, e and the golden ratio conjugate, .
That is, 2=10, 4=100, 8=1000, 16=10000
Are all zeros after a one. Perhaps this ties into the planets in terms of Si and Ge, since they make the
digital binary circuits.
log
2
3 = n
n =
log3
log2
= 1.5847
2
n
= N
P
n
= c2
n
2ϕe
(2−ϕ)
= 4.9 = Jupiter
4ϕe
(2−ϕ)
= 10 = Sat ur n
8ϕe
(2−ϕ)
= 19.69 = Ur a nus
16ϕe
(2−ϕ)
= 39.38 = Nept un e
ϕ
of 37 93
Weird Arithmetic
Since the planets seemed to defy a single expression for their distribution, I decided something weird was
going on. So, in order to find an algorithm for them, I decided to create a weird arithmetic. To do this I
decided to change the order of operations in an expression, and chose the equation of a straight line:
y=mx+b
I decided that the slope m should be 2 because for
We use the identity
Because
=
I decided that b should be -3 because the Earth is the third planet. So to do weird arithmetic I thought in
y=2x-3
That, the weird evaluation would be done by doing addition first so we take x-3 first then do
multiplication yielding
Which is
I then compared the regular arithmetic to the weird arithmetic by evaluating
In a Taylor expansion yielding:
∫
x
2
d x
1
2
+ 2
2
+ 3
2
+ … + n
2
=
n(n + 1)(2 n + 1)
6
∫
4
1
x
2
d x = lim
n→∞
n
∑
i=1
(
4i
n
)
2
4
n
lim
n→∞
64
n
3
(
n(n + 1)(2 n + 1)
6
)
=
64
3
2n − 3 → 2(n − 3)
2(n − 3) = 2n − 6
2n − 3
2n − 6
2n − 3
2n − 6
=
1
2
−
n
6
−
n
2
18
−
n
3
54
−
n
4
162
+ O(n
5
)
of 38 93
Since 1/2=0.5 is approximately Mercury orbit (0.4) I took the third term as Earth which since it
needed to be 1.00AU I chose n as which gave the orbits of the planets:
Then after the asteroids it skips to n to the sixth for Jupiter:
Thus the equation for the planets is:
Which is
i=(1, 2, 3, 4, 5..)
n
2
18
n = 18
n = 18 = 4.24264
1
2
= 0.5 = m er cur y = 0.4AU
n
6
= 0.7 = venu s = 0.72AU
n
2
18
= 1.00 = ear th = 1.00AU
n
3
54
= 1.412 ≈ 2 = m ars = 1.52AU
n
4
162
= 1.999 ≈ 2 = a ster oi d s = 2AU − 3 AU
n
6
1458
= 3.99 = jupiter = 5.2AU
n
9
39366
= 11.31AU = sa t ur n = 9.5AU
n
11
354294
= 22.624AU = ura nu s = 19AU
n
12
1062882
= 32AU = n ept u ne = 30AU
(
1
2
,
n
6
,
n
2
18
,
n
3
54
,
n
4
162
)
P
i
=
(
1
2
,
n
i
2 ⋅ 3
i
, . . .
)
of 39 93
.
.
.
Starting with Venus as . The Taylor expansion is:
Comparing regular arithmetic to weird arithmetic:
Since the equation for the distribution of the planets would the solution of a differential equation,
the equation for the distribution being:
P
1
=
n
1
2 ⋅ 3
1
=
n
6
P
2
=
n
2
2 ⋅ 3
2
=
n
2
18
P
1
∞
∑
n=0
f
n
(a)
n!
(x − a)
n
= f (a) + f′ (a)(x − a) +
f′ ′ (a)
2!
(x − a)
2
+ …
2x − 3 → 2(x − 3)
2x − 3
2x − 6
d
dx
(2x − 3)
(2x − 6)
=
f′ (x)g(x) − g′ (x)f (x)
g(x)
2
f′ (x) =
2(2x − 6) − 2(2x − 3)
(2x − 6)
2
f′ (0) = − 1/6
f′ (a)(x − a) = −
x
6
f′ ′ (a)
2!
(x − a)
2
= −
x
2
18
f′ ′ ′ (a)
3!
(x − a)
3
= −
x
3
54
f (0) =
1
2
of 40 93
Then we integrate to obtain:
Let us explore the planetary equation a little…
P
i
=
n
i
2 ⋅ 3
i
∫
n
i
2 ⋅ 3
i
di =
3
i
n
i
2log
(
n
3
)
+ C
of 41 93
We cannot write the equation of the planets as one equation but we can make a program that
outlines its logic. Here I do it in C:"
#include <stdio.h>
#include <math.h>
double r, n=4.24264;
int i;
int main(int argc, const char * argv[]) {
for (i=0;i<5;i++)
{
r=pow(n,i)/(2*(pow(3,i)));
printf("%f",r);
printf("\n");
}
int k=4;
int i=6;
while (k>0)
{
k=k-1;
r=(pow(n,i))/(2*(pow(3,i)));
i=i+k;
printf("%f", r);
printf("\n");
}
return 0;
}
of 42 93
And running it:"
Ians-MBP:~ beardsleyian$ /Users/beardsleyian/Desktop/exploring\ P_n/
weird ; exit;
0.500000
0.707107
1.000000
1.414213
1.999999
3.999996
11.313692
22.627377
31.999938
logout
Saving session...!
of 43 93
We can plot the equation 2n-3 with its weird counterpart 2n-6, which are parallel, then drop a vertical, and
a perpendicular which makes the triangle 2, , 3:
5
of 44 93
We can plot the actual planets Y1 and that of the Taylor expansion of (2n-3)/(2n-6) to get:
!
of 45 93
Another Scheme
The rate at which the distance from the sun a planet’s orbit is increases with the planets number. But as
you reach a certain critical distance from the Sun the motive force responsible for this effect stops feeling
this effect and even though position r(n) increases with respect to planetary number, n, at a certain point
the RATE of increasing, decreases. Hence the predicted value of 38.8 AU for Neptune, but the actual
value of 30 AU in this rule.
I had developed another scheme for the planets where we can look at this:
of 46 93
Amino Acids
I processed the 20 genetically encoded amino acids according to the following scheme:
In hopes of finding a connection between artificial intelligence and the biological. The result was that two
of the amino acids were equal to elements in the periodic table of the elements and they were perfectly
carbon (C) the core element of biological life, and silicon (Si) the core element of of artificial intelligence.
The amino acids are the building blocks of life, synthesized into proteins by DNA. The two amino acids
were serine and glutamine as follows,…
Ge is the other core semiconductor element.
These equations are nearly 100% accurate.
CH2+OH=12.01+2(1.01)+16,00+1.01=14.03+17.01=31.04 g/mol
Si/Ge=28.09/72.61=0.38686
a min ogr ou p
a ci dgroup
(RGroup)
H
3
N
COO
(CH
2
+ OH ) = C
H
3
N
COO
(2CH
2
+ CO + N H
2
) = Si
H
3
N
COO
= (1 − ϕ)
ϕ =
5 − 1
2
(1 − ϕ) =
Si
G e
Si
G e
(CH
2
+ OH ) = C
(CH
2
+ OH ) =
C
Si
G e
Si
G e
(2CH
2
+ CO + N H
2
) = Si
(2CH
2
+ CO + N H
2
) = G e
(CH
2
+ OH )
(2CH
2
+ CO + N H
2
)
=
C
Si
C =
(CH
2
+ OH )
(2CH
2
+ CO + N H
2
)
Si
of 47 93
(0.38686)(31.04)=12.008
C=12.01
%
2CH2=2(12.01+2.02)=28.06
CO=12.01+16.00=28.01
NH2=14.01+2.02=16.03
28.06+28.01+16.03=72.1
Ge=72.61
%
The idea is to try to understand biological life, in particular its origins, by looking at something we
understand, artificial intelligence.
The equations imply:
The primordial compounds from which amino acids are made — water (H2O) methane (CH4) and
ammonia (NH3)—seem to be related to primitive AI which would be a tungsten filament (W) encased in a
glass tube (SiO2) to make vacuum tubes for switches as follows:
12.008
12.01
100 = 99.98
72.1
72.61
100 = 99.2976
H
3
N
COO
G e ≈ Si
W
SiO
2
≈
H
2
O
CH
4
+
NH
3
CH
4
+ 1
of 48 93
Asymmetry in AI Elements
The primary elements of artificial intelligence (AI) used to make diodes and transistors, silicon (Si) and
germanium (Ge) doped with boron (B) and phosphorus (P) or gallium (Ga) and arsenic (As) have an
asymmetry due to boron. Silicon and germanium are in group 14 like carbon (C) and as such have 4
valence electrons. Thus to have positive type silicon and germanium, they need doping agents from group
13 (three valence electrons) like boron and gallium, and to have negative type silicon and germanium they
need doping agents from group 15 like phosphorus and arsenic. But where gallium and arsenic are in the
same period as germanium, boron is in a different period than silicon (period 2) while phosphorus is not
(period 3). Thus aluminum (Al) is in boron’s place. This results in an interesting equation.
The differential across germanium crossed with silicon plus the differential across silicon crossed with
germanium normalized by the product between silicon and germanium is equal to the boron divided by
the average between the germanium and the silicon. The equation has nearly 100% accuracy:
We found (Beardsley, Mathematical Structure, 2020) that the differential across silicon (P-Al) times
germanium (Ge) over boron (B) plus the differential across germanium (As-Ga) times silicon (Si) over
boron (B) was equal to the harmonic mean between Si and Ge. This was interesting because aluminum is
used as what I called a dummy doping agent element, which when inserted predicts the actually doping
agent boron, that seems out of place in the periodic table where the core artificial intelligence elements are
concerned. This is written:
I find it interesting that the AI elements can be pulled out in a 3X3 matrix because the AI elements seem
to be core to everything in the Universe. I wrote earlier:
00002: It may be we can only understand biological life relative to some other construct, like artificial
intelligence.
Stokes Theorem states:
Si(A s − G a) + G e(P − Al )
SiG e
=
2B
G e + Si
28.09(74.92 − 69.72) + 72.61(30.97 − 26.98)
(28.09)(72.61)
=
2(10.81)
(72.61 + 28.09)
0.213658912 = 0.21469712
0.213658912
0.21469712
= 0.995
Si
B
(As − G a) +
G e
B
(P − Al ) =
2 SiGe
Si + G e
∫
S
( ∇ × u ) ⋅ d S =
∮
C
u ⋅ d r
of 49 93
We know the harmonic mean H of a function is
And, that the arithmetic mean A of a function is
We have
∇ × u =
i
j k
∂
∂x
∂
∂ y
∂
∂z
u
1
u
2
u
3
i
j
k
∂
∂x
∂
∂ y
∂
∂z
0
Si
B
(Ga)z
Si
B
(As)y
=
Si
B
(As − Ga)
i
i
j
k
∂
∂x
∂
∂ y
∂
∂z
Ge
B
(Al )z 0
Ge
B
(P)x
=
Ge
B
(P − Al )
j
u = (u
1
, u
2
, u
3
)
v = (v
1
, v
2
, v
3
)
u = 0
i +
Si
B
(Ga)z
j +
Si
B
(As)y
k
u = 0
i +
Si
B
(Ga)z
j +
Si
B
(As)y
k
H =
1
1
b − a
∫
b
a
f (x)
−1
d x
A =
1
b − a
∫
b
a
f (x)d x
Si
B
(As − Ga) +
Ge
B
(P − Al ) =
Ge − Si
Ge
Si
dx
x
of 50 93
But, we want to use Stokes theorem so we want the integral in the numerator. So, we make the
approximation
And, we have
But, this is only 80% accurate. We find it is very accurate if we say
Which yields
We have by molar mass
Thus,…
We can break up our integral into two integrals u, and v:
H ≈ A
∫
1
0
∫
1
0
[
Si
B
(As − Ga) +
Ge
B
(P − Al )
]
d x d y ≈
1
Ge − Si
∫
Ge
Si
x d x
f (x) =
4
5
x
∫
1
0
∫
1
0
[
Si
B
(As − Ga) +
Ge
B
(P − Al )
]
d x d y ≈
1
Ge − Si
4
5
∫
Ge
Si
x d x
Si
B
(Ga) =
28.09
10.81
(69.72) = 181.1688g /m ol
Ge
B
(Al ) =
72.61
10,81
(26.98) = 181.2227g /m ol
Si
B
(As) =
28.09
10.81
(74.92) = 194.68111g /m ol
Ge
B
=
72.61
10.81
(30.97) = 208.02328g /m ol
u = 181z
j + 195y
k
v = 181z
i + 208y
k
of 51 93
Holding x constant and y constant in our u and v integrals we have subtle eclipses in the y-z
planes and x-z planes respectively:
By Molar Mass
We found that the differential across silicon (P-Al) times germanium (Ge) over boron (B) plus the
differential across germanium (As-Ga) times silicon (Si) over boron (B) was equal to the harmonic mean
between Si and Ge. This was interesting because aluminum is used as what I called a dummy doping
agent element, which when inserted predicts the actually doping agent boron, that seems out of place in
the periodic table where the core artificial intelligence elements are concerned. This is written:
By Density
We ask if the asymmetry in the AI elements in the periodic table due to boron results in a
dynamic equation by molar mass, then does it as well by density? While molar mass is due to the
composition of elements, density is due to the balance between the strong nuclear force holding
protons together balanced by their electric forces that are mutually repulsive.
The density of boron is 2.340 grams per cubic centimeter, that of phosphorus (white
phosphorus) is 1.88 grams per cubic centimeter and gallium is 5.904 grams per cubic centimeter.
Arsenic is 5.7 grams per cubic centimeter, germanium is 5.323 grams per cubic centimeter and
aluminum is 2.7 grams per cubic centimeter. We have the following scenario:
∫
1
0
∫
1
0
Si
B
(As − Ga)d ydz ≈
1
3
1
(Ge − Si)
∫
Ge
Si
x d x
∫
1
0
∫
1
0
Ge
B
(P − Al )d x dz ≈
2
3
1
(Ge − Si)
∫
Ge
Si
yd y
Si
B
(As − Ga) +
Ge
B
(P − Al ) =
2SiGe
Si + Ge
of 52 93
Again we see boron breaks the symmetry in that period three densities are on the same order and
period 4 densities are on the same order, but that of boron is almost the same as silicon in period
three. We see that semiconductor material Si is the the average between doping agent P and
would be doping agent aluminum that takes the place of boron and, that, the average doping
agent Ga and semiconductor material Ge is approximately the average of doping agent As. Thus
we have:
Ga-As=0.204 (differential across Ge)
Al-P=0.82 (dummy differential across Si)
Si~B
Ge/Si~B
(0.82)
Si
B
= 0.816
(0.204)
Ge
B
= 0.464
0.816
0.464
= 1.7586
GeSi = 3.52
3.52
2
= 1.76 ≈ 1.7586
(Al − P)
Ga − As
B ≈
GeSi
2
Ge
of 53 93
And, we have
The first factor takes the form of the harmonic mean between a and b, H:
And the second term takes the form of the geometric mean between a and b, G:
The equation is 94.68% accurate.
Aluminum, while a dummy in the equation used to arrive at the dynamics due to asymmetry by
way of boron actually is widely used in AI because it is a conductor, which makes it an electric
shield, so it can be used to enclose electrical circuitry to protect it from electric fields. Thus we
have the two equations by molar mass and density respectively:
But
Can be written
2 ≈
Si + P
2
B ≈
2Ge(Ga − As)
Si(Al − P) + P(Al − P)
GeSi
H =
2a b
a + b
G = ab
2(5.323)(5.904 − 5.7)
2.33(2.7 − 1.88) + 1.88(2.7 − 1.88)
(5.323)(2.33) = 2.2155
2.2155
2.340
= 94.68
B ≈
2Ge(Ga − As)
Si(Al − P) + P(Al − P)
GeSi
∫
1
0
∫
1
0
[
Si
B
(As − Ga) +
Ge
B
(P − Al )
]
d x d y ≈
1
Ge − Si
4
5
∫
Ge
Si
x d x
B ≈
2Ge(Ga − As)
Si(Al − P) + P(Al − P)
GeSi
B
2Ge(Ga − As)
[
Si(Al − P) + P(Al − P)
]
≈ G eSi
of 54 93
But
Is the geometric mean between Ge and Si. The geometric mean between a and b is given by:
Thus our equation in terms of density can be put in integral form as well:
And we see that this integral is correct:
Which is close to
94.5%
GeSi
Gf = exp
(
1
b − a
∫
b
a
log f (x)d x
)
∫
1
0
∫
1
0
[
B
2Ge(Ga − As)
[
Si(Al − P) + P(Al − P)
]
]
d x d y = exp
(
1
Ge − Si
∫
Ge
Si
log(x)d x
)
5.323(ln(5.323) − 5.323) − 2.33(ln(2.33) − 2.33) = 3.936
3.936
5.323 − 2.33
= 1.315
e
1.315
= 3.725
(5.323)(2.33) = 3.52
3.52
3.725
= 0.945
of 55 93
By Atomic Radius
of 56 93
The atomic radii data varies some from source to source. Here I use data set 4 (previous page)
that is the average of respective values in data sets 1, 2, and 3. We have looked at molar mass,
and density and the dynamic relationships that result in terms of them do to the asymmetry
introduced into the AI elements in the periodic table. It is natural to look at radius next. It has a
lot to do with the structure and properties of the elements just like is true of the their molar
masses and densities.
Here we have the differential across Si, times Si/B plus the differential across Ge times Ge/B is
the golden ratio, phi, times the arithmetic mean between Si and Ge in atomic radius.
(Al-P)=143-116=36
(Ga-As)=19
(36)(115/88)=36(1.3)=46.8
19(123/88)=19(1.4)=26.6
46.8+26.6=73.4
(115+123)/2=119
The golden ratio and the golden ratio conjugate are the solution of the quadratic
that meets the conditions and a=b+c
119
73.4
= 1.62 ≈ Φ = 1.618
(Al − P)
Si
B
+ (G a − As)
Ge
B
= Φ
Si + Ge
2
(
a
b
)
2
−
a
b
− 1 = 0
a
b
=
b
c
of 57 93
Where and , .
We have already said
Thus by radius the integral form of the equation is:
The Generalized Equation
Returning to the Asymmetry in the AI Periodic Table:
and by molar mass or
and by density or
and by atomic radius
Are respectively and
And, the ratios
and by molar mass or
and by density or
and by atomic radius
Are, quotients , and , respectively, then if
And, since the geometric mean, arithmetic mean, quadratic mean (root mean square), harmonic
mean are special cases of the generalized mean:
Φ =
a
b
ϕ =
5 − 1
2
ϕ =
1
Φ
f =
1
b − a
∫
b
a
f (x)d x
∫
1
0
∫
1
0
[
(Al − P)
Si
B
+ (G a − As)
Ge
B
]
d x d y =
Φ
Ge − Si
∫
Ge
Si
x d x
(As − Ga)
(P − Al )
(Al − P)
(Al − P)
(Al − P)
(Ga − As)
ΔE
1
ΔE
2
Si
B
Ge
B
B
2Ge(Ga − As)
Si
B
2Ge(Ga − As)
P
Si
B
Ge
B
Q
1
Q
2
∇ = (ΔE
1
, ΔE
2
)
Q = (Q
1
, Q
2
)
of 58 93
p= 0,1, 2, -1,…
Then the generalized form of our equations is:
Where C is some constant. This is all the following equations:
By molar mass.
By density.
By atomic radius.
While this is the arithmetic mean when p=1, and the harmonic mean for p=-1, I have used it as
the geometric mean for p=0, which we can see explodes to infinity at zero. But, since I am a
physicist and not a mathematician, I assume taking the limit as p—->0 is the same as evaluating
it at zero. But, as physicists undergo courses of mathematics, I do understand how to present
equations in terms of the formalities of mathematics. The proper way to treat this as a
mathematician is, the limit as p goes to zero is the geometric mean because
However we can generalize the equation to include the geometric mean without having to take
the limit as p goes to zero using the f-mean. We have:
(
1
n
n
∑
i=1
x
i
p
)
1
p
∇ ⋅ Q = C
(
1
n
n
∑
i=1
x
i
p
)
1
p
Si
B
(As − Ga) +
Ge
B
(P − Al ) =
Ge − Si
Ge
Si
dx
x
∫
1
0
∫
1
0
[
B
2Ge(Ga − As)
[
Si(Al − P) + P(Al − P)
]
]
d x d y = exp
(
1
Ge − Si
∫
Ge
Si
log(x)d x
)
∫
1
0
∫
1
0
[
(Al − P)
Si
B
+ (G a − As)
Ge
B
]
d x d y =
Φ
Ge − Si
∫
Ge
Si
x d x
M
0
(x
1
, …, x
n
) =
n
n
∏
i=1
x
i
M
f
(x
1
, …x
n
) = f
−1
(
1
n
⋅
n
∑
i=1
f (x
i
)
)
of 59 93
The power mean is obtained by letting
Thus, our equation becomes:
It is the geometric mean if
Fundamental Equations
The golden ratio and the golden ratio conjugate are the solution of the quadratic
that meets the conditions and a=b+c
Where and , .
Silicon And Carbon
We guess that artificial intelligence (AI) has the golden ratio, or its conjugate in its means geometric,
harmonic, and arithmetic by molar mass by taking these means between doping agents phosphorus (P)
and boron (B) divided by semiconductor material silicon (Si) :
Which can be written
f (x) = x
p
∇ ⋅ Q = Cf
−1
(
1
n
⋅
n
∑
i=1
f (x
i
)
)
f (x) = log(x)
(
a
b
)
2
−
a
b
− 1 = 0
a
b
=
b
c
Φ =
a
b
ϕ =
5 − 1
2
ϕ =
1
Φ
PB
Si
=
(30.97)(10.81)
28.09
= 0.65
2PB
P + B
1
Si
=
2(30.97)(10.81)
30.97 + 10.81
1
28.09
= 0.57
0.65 + 0.57
2
= 0.61 ≈ ϕ
of 60 93
We see that the biological elements, H, N, C, O compared to the AI elements P, B, Si is the golden ratio
conjugate (phi) as well:
So we can now establish the connection between artificial intelligence and biological life:
Which can be written:
Where HNCO is isocyanic acid, the most basic organic compound. We write in the arithmetic mean:
Which is nice because we can write in the second first generation semiconductor as well (germanium) and
the doping agents gallium (Ga) and arsenic (As):
Where
Where ZnSe is zinc selenide, an intrinsic semiconductor used in AI, meaning it doesn’t require doping
agents. We now have:
Germanium And Carbon
We could begin with semiconductor germanium (Ge) and doping agents gallium (Ga) and Phosphorus (P)
and we get a similar equation:
PB(P + B) + 2PB
2(P + B )Si
≈ ϕ
C + N + O + H
P + B + Si
≈ ϕ
(P + B + Si)
PB(P + B) + 2PB
2(P + B )Si
≈ (C + N + O + H )
PB
[
P
Si
+
B
Si
+ 1
]
+
2PB
P + B
[
P
Si
+
B
Si
+ 1
]
≈ 2HCNO
[
PB +
2PB
P + B
+
P + B
2
][
P
Si
+
B
Si
+ 1
]
≈ 3H NCO
[
PB +
2PB
P + B
+
P + B
2
][
P
Si
+
B
Si
+ 1
]
≈ H NCO
[
G a
G e
+
As
G e
+ 1
]
Z n
Se
≈
[
P
Si
+
B
Si
+ 1
]
[
Ga
Ge
+
As
Ge
+ 1
]
PB
(
Z n
Se
)
+
2PB
P + B
(
Z n
Se
)
+
P + B
2
(
Z n
Se
)
≈ H NCO
of 61 93
,
In grams per mole. Then we compare these molar masses to the molar masses of the semiconductor
material Ge:
Then, take the arithmetic mean between these:
We then notice this is about the golden ratio conjugate, , which is the inverse of the golden ratio, .
. Thus, we have
1.
2.
This is considering the elements of artificial intelligence (AI) Ga, P, Ge, Si. Since we want to find the
connection of artificial intelligence to biological life, we compare these to the biological elements most
abundant by mass carbon (C), hydrogen (H), nitrogen (N), oxygen (O), phosphorus (P), sulfur (S). We
write these CHNOPS (C+H+N+O+P+S) and find:
A similar thing can be done with germanium, Ge, and gallium, Ga, and arsenic, As, this time using
CHNOPS the most abundant biological elements by mass:
2G a P
G a + P
= 42.866
G a P = 46.46749
2G a P
G a + P
1
G e
=
42.866
72.61
= 0.59
G a P
1
G e
=
46.46749
72.61
= 0.64
0.59 + 0.64
2
= 0.615
ϕ
Φ
ϕ ≈
1
Φ
G a P(G a + P ) + 2G a P
2(G a + P )G e
≈ ϕ
G a P(G a + P ) + 2G a P
2(G a + P )Si
≈ Φ
CH NOPS
G a + As + G e
≈
1
2
[
G a A s +
2G a A s
G a + As
+
G a + As
2
][
G a
G e
+
As
G e
+ 1
]
≈ CHNOPS
[
G a
Si
+
As
Si
+ 1
]
G a A s
(
O
S
)
+
2G a A s
G a + As
(
O
S
)
+
G a + As
2
(
O
S
)
≈ CHNOPS
O
S
≈
[
Ga
Ge
+
As
Ge
+ 1
]
[
Ga
Si
+
As
Si
+ 1
]
of 62 93
We can also make a construct for silicon doped with gallium and phosphorus:
And for germanium doped with gallium and phosphorus:
Now we develop a scheme for naming the AI elements with numbers that might reveal patterns.
G a A s(G a + A s) + 2G a A s
2(G a + As)Ge
≈ 1
C + H + N + O + P + S
G a + As + G e
≈
1
2
(C + N + O + H ) ≈
2(G a + P )Si
G a P(G a + P ) + 2G a P
(P + B + Si)
HNCO ≈
2(G a + P )Si
(G a + P )
[
G a P +
2GaP
Ga + P
]
(P + B + Si)
HNCO ≈
2(P + B + Si)Si
G a P +
2GaP
Ga + P
G a P(G a + P ) + 2G a P
2(G a + P )G e
≈ ϕ
[
G a P +
2G a P
G a + P
+
G a + P
2
][
P
G e
+
B
G e
+
Si
G e
]
≈ H NCO
[
G a
G e
+
As
G e
+ 1
]
G a P
(
B
S
)
+
2G a P
G a + P
(
B
S
)
+
G a + P
2
(
B
S
)
≈ H NCO
of 63 93
The Fundamental AI Bioequations
[
PB +
2PB
P + B
+
P + B
2
][
P
Si
+
B
Si
+ 1
]
≈ H NCO
[
G a
G e
+
As
G e
+ 1
]
[
G a A s +
2G a A s
G a + As
+
G a + As
2
][
G a
G e
+
As
G e
+ 1
]
≈ CHNOPS
[
G a
Si
+
As
Si
+ 1
]
[
G a P +
2G a P
G a + P
+
G a + P
2
][
P
G e
+
B
G e
+
Si
G e
]
≈ H NCO
[
G a
G e
+
As
G e
+ 1
]
HNCO ≈
2(P + B + Si)Si
G a P +
2GaP
Ga + P
PB(P + B) + 2PB
2(P + B )Si
≈ ϕ
G a A s(G a + A s) + 2G a A s
2(G a + As)Ge
≈ 1
G a P(G a + P ) + 2G a P
2(G a + P )G e
≈ ϕ
G a P(G a + P ) + 2G a P
2(G a + P )Si
≈ Φ
C + N + O + H
P + B + Si
≈ ϕ
C + H + N + O + P + S
G a + As + G e
≈
1
2
Z n
Se
≈
[
P
Si
+
B
Si
+ 1
]
[
Ga
Ge
+
As
Ge
+ 1
]
O
S
≈
[
Ga
Ge
+
As
Ge
+ 1
]
[
Ga
Si
+
As
Si
+ 1
]
of 64 93
Formulating A Notation
Silicon and germanium are in group 14 like carbon (C) and as such have 4 valence electrons. Thus to have
positive type silicon and germanium, they need doping agents from group 13 (three valence electrons)
like boron and gallium, and to have negative type silicon and germanium they need doping agents from
group 15 like phosphorus and arsenic.
We need a scheme that takes all of this into account.We begin with the periodic table of the elements:
And then pull out that section with the primary artificial intelligence (AI) elements:
of 65 93
And we next number the elements as such,…
Thus 13 is an element in row 1 and must be boron because it has three valence electrons. 14 is carbon
because it is in row 1 and has four valence electrons. And so on. So, instead of element Si we have E_24,
and instead of element Ge, we have E_34, and so on. Thus,…
Becomes…
of 66 93
Where E means “element”. Now we define operators for the geometry, harmonic, and arithmetic means:
And our equation: Becomes:
And
Becomes
We can superscript E with M for molar mass, or p for density, or R for radius of the elements.
E
13
E
14
E
15
E
23
E
24
E
25
E
33
E
34
E
35
G = a b
H =
2a b
a + b
A =
a + b
2
PB(P + B) + 2PB
2(P + B )Si
≈ ϕ
G (E
25
, E
13
) +
H(E
25
, E
13
)
2E
24
≈ ϕ
[
PB +
2PB
P + B
+
P + B
2
][
P
Si
+
B
Si
+ 1
]
≈ H NCO
[
G a
G e
+
As
G e
+ 1
]
[
G (E
25
, E
13
) +
H(E
25
, E
13
) +
A(E
25
, E
13
)
]
[
E
25
E
24
+
E
13
E
24
+ 1
]
≈ H NCO
[
E
33
E
34
+
E
35
E
34
+ 1
]
of 67 93
Perfect Equations
I made two equation sets for the planetary orbits one in terms of core AI semiconducting elements Si and
Ge and one in terms of the mathematical constants and Euler’s number e. The result was that for the
first set of equations the equation for Venus was perfect and for the second set of equations the equation
for Mars was perfect. These are:
1AU= average earth-sun separation. This is interesting because Venus and Mars have always been of
great interest to space programs: they are solid and the closest to us. The Russians focused their space
program on Venus sending several probes there, presumably because it is our sister planet (close to the
same as the Earth in size and mass) to understand why she underwent a runaway greenhouse effect
presumably so we can understand how to prevent one here. And the United States has sent several rovers
to Mars to search for life and to better understand the planet in our solar system that can be colonized.
Interestingly, in 2020 we have discovered evidence of microbial life in the Venus atmosphere.
If we want to explore the geometry of the Mars equation we can express geometrically, but how do we
do as such for Euler’s number e? I propose by looking at because is the ratio between the side of
an equilateral triangle to its radius and is while (e=2.718…). Also, in the Venus
equation we have venus=0.72 AU. Thus all seems to come together with this approach. Thus we have:
Where denotes Mars as the fourth planet. We have:
Using P4=1.52, phi=0.618, Ge=72.64, and Si=28.09 we have:
ϕ
ve n us =
1
G e
2
2 SiGe +
Si
3
Ge
1 +
Si
2
Ge
2
= 0.72AU
m ars = ϕ
2
e
2−ϕ
= 1.52AU
ϕ
3
1
3
3 ≈ 1.72
e ≈ 2.72
1
G e
2
2 SiGe +
Si
3
Ge
1 +
Si
2
Ge
2
≈ ( 3 − 1)AU
ϕ
2
(1 + 3)
2−ϕ
= P
4
P
4
ln
[
P
4
ϕ
2
]
= (2 − ϕ)l n
2 Si
G e
(
Si
2
Ge
2
+ 1
)
+
Si
3
G e
3
(
Si
2
Ge
2
+ 1
)
+ 2
of 68 93
This gives:
1.38=1.2967 is
Is about 94% accuracy. We see that the common geometry to Mars and Venus is . Square root three is
irrational like in , and . This is what lends them their properties. This is clear in my illustration
on the next page of the Vesica Piscis. The interesting thing about the square root of two is that it has the
property:
One is to the square root of two as the square root of 2 is to two.
ln(3.9798) = 1.382l n
(
56.18
72.64(1.495)
+
22164.36
383290.0(1.495)
+ 2
)
1.2967
1.38
100 = 0.9396
3
5
ϕ
2
1
2
=
2
2
of 69 93
of 70 93
Here we see the relationships between the orbits of Venus, Earth, and Mars.
The Ratios Come Together
Let us say a/b=x, the golden ratio. Then,…
Let us differentiate this implicitly:
ϕ =
b
a
=
5 − 1
2
x
2
− x − 1 = 0
d
dx
x
2
−
d
dx
x −
d
dx
1 = 0
2x − 1 = 0
x =
1
2
of 71 93
Which is similar to Euler’s number, e because it is the base such that is itself :
But
Which says for this angle the x-component equals the y-component is that is , x=1/2 bisects a right
angle. Which similar in concept to Euler’s number e because it is the base such that is itself . But
if , then:
It is the diagonal of the unit square. We notice something interesting happens:
, ,
Where is the cosine of 30 degrees, in the unit equilateral triangle in which the altitude has been drawn
in (fig 14):
Since air is 25% N2 and 75% O2, the molar mass of air as a mixture is 29.0 g/mol. I found air over H2O
is the golden ratio, and in total that:
d
dx
e
x
e
x
d
dx
e
x
= e
x
sin 45
∘
= cos45
∘
=
2
2
1
2
90
∘
d
dx
e
x
e
x
sin 45
∘
= cos45
∘
=
2
2
2cos
π
4
= 2
2cos
π
n
=
2cos
π
4
= 2
2cos
π
5
= Φ
2cos
π
6
= 3
3
air
H
2
O
≈ Φ
0.75N
2
+ 0.25O
2
≈ a ir
of 72 93
Where ZnSe is zinc selenide, an intrinsic semiconductor, intrinsic meaning that it does not have to be
doped to semiconduct.
The Damped Harmonic Oscillator
The equation of a damped harmonic oscillator is:
That is to say that the force of gravity F pulling down on a mass m is equal to minus the restoring force of
the spring which is proportional to the displacement x times k the spring constant, minus the velocity with
which it slows dx/dt, times the damping constant c. This can be written:
−2cos
(
π
4
)
+ 2cos
(
π
5
)
+ 2cos
(
π
6
)(
Z n
Se
)
≈
air
H
2
O
2cos
(
π
4
)
= 2, 2cos
(
π
5
)
= Φ, 2cos
(
π
6
)
= 3
F = − k x − c
dx
dt
= m
d
2
x
dt
2
of 73 93
This has solution:
The object starts at maximum amplitude to and we want from everything that has been said that
this is . Thus so that we want . But,…
We have:
This gives us that
Thus:
Thus with incremental units of one second we actually get our constants for successive amplitudes of the
oscillator. Why do I say this? We can think of a damped pendulum as a damped harmonic oscillator, each
successive amplitude diminished by the friction of air. The period of a pendulum is:
Where l is the length of the pendulum and g is gravitational acceleration 9.8 meters per second squared.
The physicist Christaan Huygens noticed that
Which means:
If l is one meter, then the period T of the pendulum is 2 seconds. That is why he suggested that we define
the meter in terms of a pendulum that has a period of 2 seconds. We are working with one second, but that
is the time of a swing from left to right where the period T is a swing from left to right and back to left.
d
2
x
dt
2
+ 2ζω
dx
dt
+ ω
2
x = 0
x (t) = e
−ζt
2Acos(ϕ + ωt)
ϕ = 0
3
cos(ωt) = 1
ω = 2π
ω =
k
m
x (0) = e
−ζ(0)
3cos(0 + ω(0)) = 3 = 1.732
x (1) = e
−ζ(1)
3cos(0 + 2π (1)) = Φ = 1.618
ζ = 0.0681
x (2) = e
−ζ(2)
3cos(0 + 2π (2)) = 1.5 ≈ 2
T = 2π
l
g
g ≈ π
2
= 9.87
T = 2
π
2
l
g
= 2
gl
g
= 2 l
of 74 93
We have the following for our damped harmonic oscillator:
The questions arise what kind of spring constant k are we looking at, and what kind of mass m. What kind
of damping constant c are we looking at, is it given by the friction of air, or water? Or can we look at the
world of atoms and quanta, and use the wave equation:
And we say:
Light can behave like a particle and its energy, E, is given by
And that particles, like electrons, can behave like a wave given by
Where h is a constant called Planck’s constant and is given by
The energy E, above in a quantum mechanical system is analogous to the potential energy in the damped
harmonic oscillator which is even by:
Our quantum mechanical wave equation is of the form:
k
m
= 4π
2
= 39.4784
ζ =
c
2 m k
k = (39.478)m
c
m
= 8.55765 = f = ω /2π
(
∂
2
∂x
−
1
c
2
∂
2
∂t
)
ψ (x, t) = 0
E =
h c
λ
λ =
h
mv
h = 2π ℏ = 6.626 × 10
−34
J ⋅ s
U =
1
2
k x
2
U(x, t) = A
0
e
i(k x−ωt)
of 75 93
But we need to put it in terms of energy E, and momentum, p, which since"
"
"
"
"
We have the wave for our particle is"
"
QM and Electrostatics
Is the electric field of a point charge a distance r away.
Is the force a charge feels in an electric field E.
The charge of an electron equals the charge of a proton only with a reversal in sign, is:
The wavelength (lambda) of a hydrogen atom is given by the electron orbit numbers (Rydberg formula):
Is the Rydberg constant is the most accurately determined constant in physics. We have that:
(Planck’s equation)
p = ℏk
E = ℏω
k =
mv
2
2
mv
2
2
=
p
2
2m
ψ (x, t) = Ce
i
ℏ
( px−Et)
E = k
q
r
2
F = qE
k = 8.987551 × 10
9
kg ⋅ m
3
⋅ s
−2
⋅ C
−2
1.6 × 10
−19
C
1
λ
= R
H
(
1
n
2
−
1
n
1
)
R
H
= 1.09678 × 10
7
m
−1
E = h υ
λ =
c
v
c = 2.9979E17n m /s
of 76 93
Where E here is not electric field, but energy. For the force between two charges:
(Coulomb’s law)
Epsilon nought in called vacuum permittivity. It is in farads per meter.
There are four visible lines in the Balmer series given by the Rydberg formula:
v=7.31E14, 6.91E14, 6.1685E14, 4.57E14
Which gives E=4.84E-19J, 4.5786E-19J4.0866E-19J, 3.0276E-19J
If the energy in an electron is given by the first line in the Balmer series E=4.84E-19 Joules and the force
between this electron and the proton it orbits in a hydrogen atom is as if given by a spring in our harmonic
oscillator, then since:
We can find k:
From
Or,…
F = k
q
1
q
2
r
2
k =
1
4π ϵ
0
ϵ
0
= 8.854 × 10
−12
F ⋅ m
−1
U =
1
2
k x
2
E = k
q
r
2
r
2
= x
2
k =
2U
x
2
U = E
E = h υ
k =
2E
r
2
k = 2
hυ
r
2
υ =
c
λ
c = 2.9979E17n m /s
of 77 93
Note: h here is h not h bar ( ):
The average separation between an electron and a proton is giving the force between
them as:
If we imagine a light, tendril-like spring between them holding them together that has a little give so that
it can account for the variable distances of the electron from the proton with energy changes then we can
calculate the spring constant k of this hypothetical spring using the first line in the Balmer series:
That is actually a pretty strong spring constant considering it only holds together something as light as an
electron and a proton (note that it is in kilograms not grams!). Let us compare this with a spring that has a
spring constant that produces a force of earth gravity (9.81 meters per second squared) for a displacement
of one meter from our interesting pendulum example:
x=1m,
Thus a spring with Earth gravity that will move one kilogram of mass through a distance of one meter at
earth acceleration has a spring constant of 9.81 kilograms per second squared. The proton spring is 35
times stronger:
That would explain why matter is so strongly held together. The Balmer series has four emission lines in
the visible spectrum. We used the first one in this calculation which corresponds to the lowest energy state
of an electron. When an electron absorbs energy it jumps to an outer orbital of a higher energy state.
Eventually it falls back into to the lower orbital emitting energy proportional to to the change: this is the
emission of photons of light of energy given by . One comparison that people often make is the
strength of gravity compared to the strength of these electric forces of a proton holding an electron. They
do it by comparing the gravity of a proton due to its mass to the electric force it produces. I see this as
erroneous because while the gravity is due to the mass of the proton, I don’t think we know that the
electric force the proton causes is due to the mass of the proton. However, I do think we can compare the
strength of gravity to the strength of electric forces in a proton as done here with springs. Let us look at
ℏ
h = 2π ℏ = 6.626 × 10
−34
J ⋅ s
0.530 × 10
−10
m
F = 8.99 × 10
9
(1.6 × 10
−19
)(1.6 × 10
−19
)
(0.530 × 10
−10
)
2
= 8.19 × 10
−8
N
k = 2
4.84 × 10
−19
(0.530 × 10
−10
)
2
= 344.6kg /s
2
F = − k x
a = 9.81m /s
2
(1kg)(9.81m /s
2
) = − k (1m)
k = − 9.81k g /s
2
344.6
9.81
= 35
E = h υ
of 78 93
the other method of using the mass of a proton. We already calculated that the Force between two charges
is:
Thus, using Newton’s Universal Law of Gravity we have:
This gives:
Meaning giving electric forces billions of times stronger than gravity! I
merely had 35!
Admitedly I am using the gravity of the earth which is not the only gravity that exists, but I think in order
to compare Electric forces due to charge q to gravitational forces due to mass m, we would have to
express them in terms of some common denominator like a spring model. If we want to know the spring
constants for these respective forces, it may be we can find them in their respective constants of
proportionality.
The Planetary Equations
Though I did not derive these equations, but guessed at them, it was an educated guess which proceeded
from the argument, that the first planet being the closest to the Sun, sets the idea in motion, that idea being
that its distance is in the simplest expression between Si, and Ge, possible; the ratio between them. Thus
for Mercury ( ) we have
In astronomical units because we take an astronomical unit to be 1 AU at earth orbit, because the earth is
the one planets in the solar system that is highly hospitable to life.
To make our guess at the distance of the next planet, we first guess at the simplest idea r(n)=n. Seeing this
does not work we go the next simplest expression r(n)=2n. Seeing this does not work we go one step
higher in complexity and try . We see this does not work either so we go one step of complexity
beyond that and we try . And, we find this works. But it must be tempered with a factor of 0.3
and and adjusted by 0.4. Thus we have the Titius-Bode Rule for the distribution of the planets:
F = 8.99 × 10
9
(1.6 × 10
−19
)(1.6 × 10
−19
)
(0.530 × 10
−10
)
2
= 8.19 × 10
−8
N
F = G
m
1
m
2
r
2
F = (6.67 × 10
−11
)
(9.11 × 10
−31
kg)(1.67 × 10
−27
kg)
(0.530 × 10
−10
m)
2
= 3.61 × 10
−47
N
8.19 × 10
−8
3.61 × 10
−47
= 2.27 × 10
39
P
1
P
1
=
Si
Ge
r (n) = n
2
r (n) = 2
n
r (n) = 0.4 + (0.3)2
n
n = − ∞, 0,1, 2,…
of 79 93
To get the next equation for . Since we are dealing with a doubling effect we guess it is ,
or . Which is close but a little too high. So we guess at something lower and that it involves twice
their product and something big, like the sum of their squares to reduce the number to prevent the product
from being too large. We guess that that value then is . This is a little too low so we average
the two to get exactly the result we need:
Since the next planet, our Earth, must be at a greater value than Venus, and the uppercase scenario for
Venus is , we want to reduce 2SiGe by an amount less than the sum of their squares, we
reduce it by the difference of their squares and get for Earth:
It takes a bit of doing, but the next planet ( ) is:
The next location is the asteroids from 2.2 AU-3.2 AU. Since Mars is the last terrestrial planet (solid), and
the asteroid is a bunch of rocks that could not form into a solid planet and after the asteroids we have the
gas giants Jupiter, Saturn, Uranus, Neptune…and this represents a flipping around point around the
asteroid belt, I started counting again with a new pattern starting with Jupiter as , and flipped the Earth
equation and turned a minus sign to a plus to obtain the Jupiter equation, which is a quadratic in its
simplest form in the numerator, and a product in its simplest form in the denominator, which is great
because it is the first planet after the asteroid belt and is hence:
Now what happens is that a simple pattern forms. We simply multiply Jupiter by 2 to get Saturn, then by
four to get Uranus, and finally by six to get Neptune. We have:
P
2
= Venu s
2P
1
2
Si
Ge
2SiG e
Si
2
+ G e
2
P
2
=
1
Ge
2
2SiG e +
Si
3
Ge
1 +
Si
2
Ge
2
2SiG e
Si
2
+ G e
2
P
3
=
2SiG e
Ge
2
− Si
2
P
4
= Mars
P
4
=
2SiG e
2
(Si − Ge)
2
(Si + G e)
P
1
P
1
=
Si
2
+ 2SiG e + G e
2
SiG e
P
2
= Sat ur n =
2(Si + G e
2
SiG e
P
3
= Ura nu s =
4(Si + G e
2
SiG e
P
4
= Nept u n e =
6(Si + G e
2
SiG e
of 80 93
The Geometry
Also, to look at life as a mathematical construct, look at CH4 which is one of the primordial earth
substances (methane) and is part of making amino acids, the building blocks of life. Since H is H1+ and C
is C4-, then to have neutral methane:
To look at this mathematically (as a mathematical construct) think of it in terms of geometry: Since C is
C4- meaning it has four valence electrons, think of it as a regular fours sided shape (a square) and H as a
regular octagon or truncated square:
If each side of the carbon square is one, then the radius r1 is square root 2 over 2:
of 81 93
Since we are saying hydrogen is a regular octagon then its r, r2 is the quantity one plus square root of 2
over 2:
of 82 93
Which takes us to the square root of three. Let us consider the other primordial substance that was in the
early earth that gave rise to amino acids, the building blocks of life, ammonia (NH3). Since N (nitrogen)
is N3- and hydrogen (H) is H1+, we have:
If the side is s=1, then the ratio of the side s to the radius r3 is:
Is our square root of three from considering methane (CH4). Thus, where carbon (C4-) represents (is
represented by) a square, then ammonia (NH3) has nitrogen (N3-) represented by an equilateral triangle.
The other pivotal substance that was in the early earth used to make amino acids, the building blocks of
life, was water (H2O). Since H is H1+ and O is O2- we have:
1
1
3
=
s
r
3
= 3
of 83 93
Comparing the Elements Within a Compound
Here instead of comparing one compound to another, we compare the elements in the compound to one
another and the compound itself. We start with bone, for which as we have said before, the mineral
component is hydroxyapatite (HA) and the organic component is collagen. Thus we have:
Calcium (Ca)=40.078 g/mol
Phospate anion (PO4)=30.973 g/mol
Ca5=200.39 g/mol
(PO)3=92.919g/mol
Hydroxide ion (OH)=17.01g/mol
C=12.01 g/mol
H=1.01 g/mol
N=14.01g/mol
O=16.00 g/mol
We begin comparing:
….
H A = Ca
5
(PO
4
)
3
OH
1.55g /c m
3
wh ite = 1.823g /c m
3
red = 2.2 − 2.34g /c m
3
violet = 2.36g /c m
3
bl a ck = 2.69g /c m
3
Coll agen = C
57
H
91
N
19
O
16
Ca
5
Ca
5
(PO
4
)
3
OH
=
200.39
502.32
= 0.3989 ≈ 0.4
(PO
4
)
3
Ca
5
(PO
4
)
3
OH
=
92.919
502.32
= 0.18497969 ≈ 0.2
OH
Ca
5
(PO
4
)
3
OH
=
17.01
502.32
= 0.00199 ≈ 0.002
of 84 93
Calcium Ca and carbon C are the primary mineral components of the respective compounds HA and
Collagen. Thus, we call them the i vector. Phosphate and nitrogen are the primary salt components, thus
we call them the j vector. OH and O are the oxygen components, therefore we call them the k vectors. We
have:
Is the angle between the HA and Collagen vectors. Taking the cross product we have:
Which has a magnitude 0.0906. We already know that AB=0.25678 and since:
We have that:
Which is a theta that is around 21 degrees, which we see is right because we got the same value for the
dot product. But just what are we doing when we cross g/mol with g/mol? We look at that:
C
57
C
57
H
91
N
19
O
16
=
684.57
1302.5
= 0.52558 ≈ 0.5
H
91
C
57
H
91
N
19
O
16
=
91.91
1302.5
= 0.07
N
19
C
57
H
91
N
19
O
16
=
266.19
1302.5
= 0.204 ≈ 0.2
O
16
C
57
H
91
N
19
O
16
=
256
1302.5
= 0.1965 ≈ 0.2
A ⋅
B = (Ca
5
)(C
57
) + (PO
4
)
3
N
19
+ (OH )(O
16
) = (0.4)(0.5) + (0.2)(0.2) + (0.002)(0.2) = 0.24
A B = 0.4
2
+ 0.2
2
+ 0.002
2
0.5
2
+ 0.2
2
+ 0.2
2
= 0.25678
A ⋅
B = A Bcosθ
θ = 20.83
∘
i j k
Ca
5
(PO
4
)
3
OH
C
57
N
19
O
16
= 0.0396i − 0.079j − 0.02k
A × B = A Bsi n θ
0.0906 = 0.25678sin θ
sinθ =
0.0906
0.25678
θ = 20.66
∘
of 85 93
It is grams squared per moles squared acting on grams per mole through an angle, which would be like
torque in our instance of grams per mole of the mineral component on the organic matrix through which
it is distributed.
The mineral component of bone is the solid, hard atoms of bone which are embedded in the organic
collagen matrix. So what we are doing is analyzing the forces over which the HA does not tear apart the
collagen. The organic collagen matrix through which the the HA is distributed. So, at this point we might
want to look at the density of bone and the density of collagen to find if the mathematical structure we
have found predicts the experimental strength of bone. If we look at the HA and collagen in such a way
that we get cancellation of grams and end up with moles and cubic centimeters we will be looking at:
But really the moles squared per grams squared is a normalized molar mass, the molar mass in terms of
the molar masses of the compounds of which they are a part. Really we are dealing with unitless numbers.
The density of HA is:
For dry collagen it ranges from 0.0076 and 0.0311 with pore sizes between 250 and 350 a
range in which it scaffolds for functional soft tissue growth. Since the upper value is for strongest bone,
we will use it. We have:
We now divide this by our A dot B which is in grams squared per moles squared:
Now we take the cube root:
We now find if we take the arc sine or arc cosine of this we have one equals the other is 45 degrees.
A ×
B = (g . m ol )(g /m ol )
i + (g /m ol )(g /m ol )
j + (g /m ol(g /m ol )
k
A ×
B = (g
2
/m ol
2
)
i + (g
2
/m ol
2
)
j + (g
2
/m ol
2
)
k
A ×
B = (g
2
/m ol
2
)
2
+ (g
2
/m ol
2
)
2
+ (g
2
/m ol
2
)
2
= g
2
/m ol
2
τ
m ol
2
g
2
⋅
g
cm
3
=
m oles
2
gr a m s ⋅ cm
3
H A = 3.00g /c m
3
g
cm
−3
μm
m ol
2
0.0906g
2
0.0076g
cm
3
= 0.084
m ol
2
g ⋅ c m
3
0.084
m ol
2
g ⋅ c m
3
m ol
2
0.24g
2
= 0.35
m ol
4
g
3
⋅ cm
3
3
0.35 = 0.7047
sin
−1
(0.7047) = 44.8
∘
≈ cos
−1
(0.7047) = 45.19
∘
≈ 45
∘
of 86 93
This is the angle in the diagonal of a unit square (see fig. 14).
This value:
The thing to do here is to do the calculation with dimensionless values.
The units are:
Which makes sense: grams usually occupy space, grams per centimeter cubed. But here, we have not
centimeters cubed, but centimeter to the one, a straight line. Thus if you want to have grams in a straight
line (cm) you have to have the cube root of grams, because mass is three dimensional, and the line (cm) is
one dimensional. It is not out of the ordinary to present mass like this: The thin disc approximation for the
distribution of mass over a two dimensional disc can very closely approximate the mass in its actual three
dimensional manifestation, only integrating over two dimensions. That which we are seeing here is the
cross section of three dimensional atoms over a straight line, which can accurately be represented by three
dimensional atoms because atoms are so small, distributed over a line, it is as if a three dimensional atom
(a small sphere) is like a one dimensional point.
1
0.0906
0.0076g
cm
3
= 0.084
g
cm
3
0.084
g
cm
3
1
0.24
= 0.35
g
cm
3
3
0.35 = 0.7047
3
0.35 = 0.7047
3
0.35 = 0.7047
gr a m s
1/3
cm
of 87 93
Conversion Factor
Methane (CH4) is one of the most fundamental compounds, a carbon atom (C) the core
element of biological life, surrounded by four hydrogens. It is the most fundamental organic
compound in that it bonds:"
"
We want to convert it into geometry and we find if we do it is the square and regular octagon:"
"
We see that since r, the apothem of the regular octagon is and a, the apothem of
the square is then the distance from carbon (C) of hydrogen (H) is
is approximately the square root of 3: . We know from
chemistry that the hydrogen bond is 0.11nm. It is a covalent bond meaning carbon shares its
four outer valence electrons with up to 4 hydrogen completing both their outer shells.
Computations of the geometry on the next page…!
r =
2 + 1
2
a =
1
2
a + r =
2 + 2
2
= 1.707
3 = 1.732
of 88 93
!
of 89 93
If d=1.707=0.11nm where 0.11nm is the separation between hydrogen and carbon in methane,
then the unit factor to convert nanometers to mathematical units, and mathematical units to
nanometers is 15.518 units/nm. We can use these to compute the apothem a of a square in
nano meters and the apothem r of regular octagon. We have said a=1/2=0.5 units and that the
apothem r=1.207 units yielding for the square and regular octagon respectively:"
"
"
Because ."
We know the radius of carbon is:"
"
And that of hydrogen is:"
"
Using our unit factor we can convert these to mathematical units as based on our theory for
methane being the most basic hydrocarbon unit:"
"
"
The common side s to both the square and regular hexagon are: 2(a)=2(0.5)=1 unit"
a =
nm
15.518units
⋅
0.5units
1
= 0.03222nm
r =
nm
15.518units
⋅
1.207units
b
= nm = 0.07778nm
r =
2 + 1
2
r
C
= 0.077nm
r
H
= 0.037nm
r
C
=
15.518units
nm
⋅
0.077
nm
= 1.194units
r
H
=
15.518units
nm
⋅
0.037nm
1
= 0.574units
of 90 93
Here we see how it all comes together in a 30-60-90 triangle…!
of 91 93
And we see it all together here:"
We have a conversion factor for atomic structure in terms of geometry based on methane.!
of 92 93
Conclusion
These works strive to show that life is not just chemical, but mathematical, and to establish the connection
between biological life and artificial intelligence. It is suggested we can only understand biological life
relative to some other construct, like artificial intelligence. This work established the patterns,
organization, and structure in biological life elements, artificial intelligence elements, and element
synthesis in stars.
1. It may be that whatever engineered the Universe and its elements, most or all of which occur on our
planet, it may have been as such so that they could be organized into the 18 groups with their seven
periods, such that they could be organized in such a way that it could at least enable us to build ships that
can jump through hyperspace to any star in our galaxy or beyond it. It may be that this is so because they
can be organized using mostly just four of them (C, N, O, H) in such a way that they can result in
something as complex and functional as the biological organism. It may be that the biological organism
mirrors in organization any such technology such as starships.
2. The purpose of stars is to produce not just biological elements, but artificial intelligence (AI) elements.
Because if we include AI elements in the life elements we have a mathematical pattern in the periodic
table of the elements that determines an equation that predicts the equations of nucleosynthesis in the
stars.
3. Is it the purpose of stars, as well, to produce malleable and ductile metals for humans to make tools
early on (The Iron Age) and copper wire (Age of Electronics). Or even that silicon dioxide (obsidian)
flakes into sharp edges so that we could create spearpoints with which to hunt in our beginning.
4. We are evolving towards a culminated form of specific mathematical proportions. We are evolving
towards an intermembral index of perfectly the square root of two uniformly throughout the human
species.
5. It is a purpose of biological life (C, N, O, H) to discover the properties of (P, B, Si) so it can make
computing machines which are ultimately necessary to its survival.
of 93 93
The Author
